Proof: Divide Numerators
Let's prove the following theorem:
if the following are true:
- a / c = b / c
- not (c = 0)
then a = b
Proof:
Given
| 1 | a / c = b / c |
|---|---|
| 2 | not (c = 0) |
| # | Claim | Reason |
|---|---|---|
| 1 | a / c = a ⋅ (1 / c) | a / c = a ⋅ (1 / c) |
| 2 | b / c = b ⋅ (1 / c) | b / c = b ⋅ (1 / c) |
| 3 | a ⋅ (1 / c) = b ⋅ (1 / c) | if a / c = b / c and a / c = a ⋅ (1 / c) and b / c = b ⋅ (1 / c), then a ⋅ (1 / c) = b ⋅ (1 / c) |
| 4 | (a ⋅ (1 / c)) ⋅ c = (b ⋅ (1 / c)) ⋅ c | if a ⋅ (1 / c) = b ⋅ (1 / c), then (a ⋅ (1 / c)) ⋅ c = (b ⋅ (1 / c)) ⋅ c |
| 5 | (a ⋅ (1 / c)) ⋅ c = a | if not (c = 0), then (a ⋅ (1 / c)) ⋅ c = a |
| 6 | (b ⋅ (1 / c)) ⋅ c = b | if not (c = 0), then (b ⋅ (1 / c)) ⋅ c = b |
| 7 | a = b | if (a ⋅ (1 / c)) ⋅ c = (b ⋅ (1 / c)) ⋅ c and (a ⋅ (1 / c)) ⋅ c = a and (b ⋅ (1 / c)) ⋅ c = b, then a = b |
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